Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<list>
#include<bitset>
#include<stack>
#include<memory>
using namespace std;
using namespace std::chrono;
typedef long long int llint;
typedef double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase
/*cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);*/
const llint mod=1000000007;
const llint big=2.19e18+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-15;
template <class T,class U>bool mineq(T& a,U b){if(a>b){a=b;return true;}return false;}
template <class T,class U>bool maxeq(T& a,U b){if(a<b){a=b;return true;}return false;}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){if(a==0){return b;}return a/gcd(a,b)*b;}
template<class T> void SO(T& ve){sort(ve.begin(),ve.end());}
template<class T> void REV(T& ve){reverse(ve.begin(),ve.end());}
template<class T>llint LBI(vector<T>&ar,T in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
template<class T>llint UBI(vector<T>&ar,T in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}

int main(void){
	int n,i,m,s,t;cin>>n>>m>>s>>t;s--;t--;
	vector<vector<int>>go(n);
	vector<int>sdis(n,mod);
	vector<int>tdis(n,mod);
	for(i=0;i<m;i++){
		int x,y;cin>>x>>y;x--;y--;
		go[x].pub(y);
		go[y].pub(x);
	}
	queue<int>que;
	que.push(s);
	sdis[s]=0;
	while(que.size()){
		int x=que.front();
		
		que.pop();
		for(auto it:go[x]){
			if(sdis[it]!=mod){continue;}
			sdis[it]=sdis[x]+1;
			que.push(it);
		}
	}
	//cerr<<sdis[t]<<endl;
	que.push(t);
	tdis[t]=0;
	while(que.size()){
		int x=que.front();
		que.pop();
		for(auto it:go[x]){
			if(tdis[it]!=mod){continue;}
			tdis[it]=tdis[x]+1;
			que.push(it);
		}
	}
	int f=sdis[t];
	//cerr<<f<<endl;
	vector<llint>sdn(n);
	vector<llint>tdn(n);
	for(i=0;i<n;i++){
		if(sdis[i]<n){sdn[sdis[i]]++;}
		if(tdis[i]<n){tdn[tdis[i]]++;}
	}
	llint ans=0;
	for(i=0;i+2<=f;i++){ans+=sdn[i]*tdn[f-i-2];}
	cout<<ans<<endl;
	return 0;
}